Question

calculate the cell potential ($mathrm{e_{cell}^circ}$) for the reaction given below.
$2mathrm{li}(s) + mathrm{f_2}(g) \
ightarrow 2mathrm{li^+}(aq) + 2mathrm{f^-}(aq)$

half-reaction	$mathrm{e^circ}$ (v)

| $mathrm{li^+ + e^- \
ightarrow li}$ | -3.05 |
| $mathrm{f_2 + 2e^- \
ightarrow 2f^-}$ | +2.87 |
$mathrm{e_{cell}^circ} = ?$ v

Explanation:

Step1: Identify oxidation and reduction half - reactions

In the overall reaction \(2Li(s)+F_2(g)
ightarrow2Li^+(aq) + 2F^-(aq)\), lithium is oxidized (loses electrons) and fluorine is reduced (gains electrons).

The oxidation half - reaction is the reverse of \(Li^++e^-
ightarrow Li\). So the oxidation half - reaction is \(Li(s)
ightarrow Li^+(aq)+e^-\), and the standard oxidation potential \(E_{ox}^o\) is the negative of the standard reduction potential of the reverse reaction. The standard reduction potential for \(Li^++e^-
ightarrow Li\) is \(E_{red,Li}^o=- 3.05\ V\), so \(E_{ox,Li}^o = 3.05\ V\) (since oxidation is reverse of reduction, \(E_{ox}^o=-E_{red}^o\) for the reverse reaction).

The reduction half - reaction is \(F_2(g)+2e^-
ightarrow2F^-(aq)\) with \(E_{red,F_2}^o = 2.87\ V\).

Step2: Calculate the cell potential

The formula for the standard cell potential \(E_{cell}^o\) is \(E_{cell}^o=E_{red}^o(\text{cathode})+E_{ox}^o(\text{anode})\) (or we can also use \(E_{cell}^o = E_{red}^o(\text{cathode})-E_{red}^o(\text{anode})\), where \(E_{red}^o(\text{anode})\) is the reduction potential of the oxidation half - reaction).

In the overall reaction, the cathode is where reduction occurs (fluorine reduction) and the anode is where oxidation occurs (lithium oxidation).

Using the formula \(E_{cell}^o=E_{red}^o(\text{cathode})-E_{red}^o(\text{anode})\), where \(E_{red}^o(\text{cathode}) = 2.87\ V\) (for \(F_2\) reduction) and \(E_{red}^o(\text{anode})\) is the reduction potential of the reaction at the anode (the reverse of the oxidation reaction, i.e., \(Li^++e^-
ightarrow Li\) with \(E_{red}^o=-3.05\ V\)).

So \(E_{cell}^o=2.87\ V-(- 3.05\ V)\)
\(E_{cell}^o=2.87 + 3.05\)
\(E_{cell}^o = 5.92\ V\)

Answer:

\(5.92\)